algebra exercise about finding its roots in $C$ asked by Arshad Mughal in the crazy mathematics

Exercise

Solve, ${{x}^{4}}+4=0$

 Solution:

We have ${{x}^{4}}+4={{\left( {{x}^{2}} \right)}^{2}}+{{\left( 2 \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}-{{i}^{2}}{{2}^{2}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left( 2i \right)}^{2}}=\left( {{x}^{2}}-2i \right)\left( {{x}^{2}}+2i \right)=0$

$\Rightarrow {{x}^{2}}-2i=0\,\,\,or\,\,{{x}^{2}}+2i=0$

$\Rightarrow {{x}^{2}}=2i\,\,or\,\,{{x}^{2}}=-2i$

From ${{x}^{2}}=2i$ we obtain two complex roots

Let $x=r{{e}^{i\theta }}\Leftrightarrow {{x}^{2}}={{r}^{2}}{{e}^{i2\theta }}=2i=2{{e}^{i\frac{\pi }{2}}}$

$\Rightarrow {{r}^{2}}=2\,\,\,\And \,\,2\theta =\frac{\pi }{2}+2k\pi \Rightarrow r=\sqrt{2}\,\,\,\And \,\,\theta =\frac{\pi }{4}+k\pi$ , $0\le k\le 1$

Thus $x=\sqrt{2}{{e}^{i\frac{\pi }{4}}}=\sqrt{2}\left( \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right)=1+i$  when $k=0$

And $x=\sqrt{2}{{e}^{i\frac{5\pi }{4}}}=\sqrt{2}\left( -\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \right)=-1-i$ when $k=1$

Also we get two roots from ${{x}^{2}}=-2i$ by the same procedure

Using $x=r{{e}^{i\theta }}\Leftrightarrow {{x}^{2}}={{r}^{2}}{{e}^{i2\theta }}=-2i=2\left( -i \right)=2{{e}^{i\frac{3\pi }{2}}}$

$\Rightarrow r=\sqrt{2}\,\,\And \,\,2\theta =\frac{3\pi }{2}+2k\pi \Rightarrow \theta =\frac{3\pi }{4}+k\pi \,\,\,\,,\,\,\,\,\,0\le k\le 1$

Thus $x=\sqrt{2}{{e}^{i\frac{3\pi }{4}}}=\sqrt{2}\left( -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \right)=-1+i$ when $k=0$

And $x=\sqrt{2}{{e}^{i\frac{7\pi }{4}}}=\sqrt{2}\left( \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \right)=1-i$ when $k=1$

Therefore $x\in \left\{ 1+i,-1-i,-1+i,1-i\,\,\,;\,\,i\in \mathbb{C} \right\}$